Question

integral of 1/(x^2+8x+80)
I would assume that I'm going to try and get this to equal arctan

Answer 1

\[\int\limits\frac{ 1 }{ x^2+8x+80 }\]

Answer 2

I would try and get this into something like \[\frac{ 1 }{ 1+x^2 } = \arctan\] I think

Answer 3

Try complete square for the denominator!

Answer 4

That is what I'm relearning quickly, at the moment. Haha

Answer 5

@sjerman1 Can you?

Answer 6

Ok, so you're correct that you're trying to get it into a form that is able to be integrated to an arctan equation. How you would want to do that is by completing the square on the bottom.
\[1/((x^{2}+8x+16)+64)\] =
\[1/((x+4)^{2}+64\]
Using u = x+4, and a = 8, this integral will give you

\[\frac{ 1 }{ 8 }arctan((x+4/8))\]

Answer 7

Yes, I haven't completed the square for a while now but I am getting a refresher course on Khan academy right now. :)

Answer 8

Good place for resources!

Answer 9

I suppose @JonathanJ miss the parentheses ( x + 4 )/8

Answer 10

@Chlorophyll Oh, yes I did, my bad.

Answer 11

I know just in case @sjerman1 unsure :)

Answer 12

Alright, crash course over, very easy just \[ax^2+bx+c \therefore [\frac{ b }{ 2 }]^2 = z \therefore ax^2+bx+z=-c+z \therefore (a+\sqrt{z})^2+(-c+z)\]

Answer 13

Haha, if anyone can understand that. Sooo, now i should be able to pick up with \[\frac{ 1 }{ (x+4)^2+64 }\]

Answer 14

alright @JonathanJ I understand that we are using u-sub for the x+4 and that root 64 is 8 but how did you jump to 1/8 arctan (x+4)/8 ? @Chlorophyll would you know?

Answer 15

By the formula:
∫ dx/( u² + a² )

Answer 16

Alright, I can see that from \[\int\limits \frac{ 1 }{ 1+x^2 }dx=arcsinx+C\] but how does that turn into u/a?

Answer 17

First of all, arctan , not arcsin
second, u = x + 4, a = 8, ( not 1)

Answer 18

I'm sorry I misstyped when looking at my equation sheet. but the general equation is 1/(1+x^2) = arctanx+C

Answer 19

Not always, you get simple form x² + 1
usually (ax + b)² + c²

Answer 20

Right o - but getting back to the equation how does the formula:
∫ dx/( u² + a² ) help out with a quick solve for this to become the 1/8arcsin(x+4)/8?

Answer 21

Looking at this in wolf they seem to take it down another step with second u-sub but I'm interested in this seemingly shortcut @Chlorophyll

Answer 22

Why do you keep saying arcsin ?

Answer 23

I'm soooo sorry, I am quite sleepy and keep using arcsin when I should be saying arctan... >.<

Answer 24

Suppose x + 4 is X, then a² = 1
-> the result will be: (1/ 1 )arctan X/ 1
Now plug x + 4 and a = 8 back, can you?

Answer 25

I guess that makes sense

Answer 26

@sjerman1 In case you see this, I left, sorry.

The formula for arctan is this (and how i came upon the answer)

∫1/(u^(2)+a^(2)) du = (1/a)tan^(-1)(u/a)

Answer 27

Haha funny enough @JonathanJ @Chlorophyll, we learned about this formula during class on Monday... It helped me get a jump on the new material. Thank you both.